An LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring mass damped oscillator having damping constant b, the correct equivalence of b would be:

Options

A. `b harr R`
B. `b harr c`
C. `b harr (1)/(R )`
D. `b harr (1)/(L)`

Explanation

To find the equivalence of the damping constant \( b \) in an LCR circuit compared to a physical spring-mass damped oscillator, we can follow these steps:

### Step 1: Understand the equations of motion for both systems

For a spring-mass system, the equation of motion can be written as:
\[ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 \]
where:
- \( m \) is the mass,
- \( b \) is the damping constant,
- \( k \) is the spring constant,
- \( x \) is the displacement.

### Step 2: Write the equation for the LCR circuit

In an LCR circuit, the relationship between the charge \( Q \), inductance \( L \), capacitance \( C \), and resistance \( R \) can be expressed as:
\[ L \frac{d^2Q}{dt^2} + R \frac{dQ}{dt} + \frac{Q}{C} = 0 \]
where:
- \( L \) is the inductance,
- \( R \) is the resistance,
- \( C \) is the capacitance,
- \( Q \) is the charge.

### Step 3: Compare the two equations

Now, we can compare the two equations. The equation for the spring-mass system is:
\[ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 \]
and for the LCR circuit, we have:
\[ L \frac{d^2Q}{dt^2} + R \frac{dQ}{dt} + \frac{Q}{C} = 0 \]

### Step 4: Identify the equivalent terms

From the comparison, we can see that:
- The term \( b \frac{dx}{dt} \) in the spring-mass equation corresponds to \( R \frac{dQ}{dt} \) in the LCR circuit.
- Therefore, we can conclude that the damping constant \( b \) in the spring-mass system is equivalent to the resistance \( R \) in the LCR circuit.

### Conclusion

Thus, the correct equivalence of the damping constant \( b \) in the LCR circuit is:
\[ b = R \]